# Question 12:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x^3 - 9x^2 + 26x - 18 \Rightarrow \frac{dy}{dx} = 3x^2 - 18x + 26$ | M1A1 | Two of the three terms correct (may be unsimplified). $\frac{dy}{dx} = 3x^2-18x+26$ need not be simplified |
| At $x=4 \Rightarrow \frac{dy}{dx} = 3\times4^2 - 18\times4 + 26\ (=2)$ | M1 | Substitutes $x=4$ into their $\frac{dy}{dx}$ |
| Equation of normal: $y - 6 = -\frac{1}{2}(x-4) \Rightarrow 2y+x=16$ | dM1A1* | Must have scored both M's. Look for $y-6=-\frac{dx}{dy}\bigg|_{x=4}\times(x-4)$. If form $y=mx+c$ used, for proceeding as far as $c=$.. cso $2y+x=16$, note this is a given answer |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $x=1$ in $(y) = 1-9+26-18=0$ | B1* | Either substitute $x=1$ in $(y)=1^3-9(1)^2+26(1)-18=0$ or substitute $y=0$ to reach $(x-1)(x^2-8x+18)$ by inspection or division and state $x=1$ |

## Part (c) — Way One (area under curve):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^3-9x^2+26x-18\,dx = \left[\frac{1}{4}x^4-3x^3+13x^2-18x\right]$ | M1A1 | Integrates with at least two terms correct (unsimplified) |
| Normal meets $x$ axis at $x=16$ | B1 | For sight of normal meeting $x$ axis at 16. May be embedded within formula or on diagram |
| Area of triangle $= \frac{1}{2}\times(16-4)\times6\ (=36)$ | M1 | Correct method for area of triangle. Either $\frac{1}{2}\times$('16'$-4)\times6$ or correct integration with limits 4 to their 16: $\left[8x-\frac{1}{4}x^2\right]_4^{x='16'}$ |
| Correct method for area $= \left[\frac{1}{4}x^4-3x^3+13x^2-18x\right]_1^4 + \frac{1}{2}\times(16-4)\times6$ | dM1 | Dependent on both M's. Fully correct method for area of $R$. Limits of integral(s) must be correct |
| $= 51.75$ | A1 | cso 51.75, oe such as $\frac{207}{4}$ |

## Part (c) — Way Two (area between line and curve):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrates $\left(8-\frac{1}{2}x\right)-\left(x^3-9x^2+26x-18\right)$ with at least two terms correct | M1 | Condone slips on straight line (eg $8-x$, $16-x$) or on the subtraction |
| $\pm\left\{\left(8x-\frac{1}{4}x^2\right)-\left(\frac{1}{4}x^4-3x^3+13x^2-18x\right)\right\}$ | A1 | Completely correct integration either way around |
| Normal meets $x$ axis at $x=16$ | B1 | May be implied within a triangle formula from diagram |
| $\frac{1}{2}\times$('16'$-1)\times y_{NORM\ at\ x=1} = \left(\frac{1}{2}\times\text{'16'-1}\times\text{'7.5'} = 56.25\right)$ or correct integration with limits 1 to their 16: $\left[8x-\frac{1}{4}x^2\right]_1^{x='16'}$ | M1 | Correct method for area of large triangle. Base = (their 16 $-1$). Height must be attempt to find $y$ value on normal at $x=1$ |
| Correct method for area $= \frac{1}{2}\times\text{'16'-1}\times\text{'7.5'} - \left[\left(8x-\frac{1}{4}x^2\right)-\left(\frac{1}{4}x^4-3x^3+13x^2-18x\right)\right]_1^4$ | dM1 | Dependent on both M's. Fully correct method. Limits of integral(s) must be correct |
| $= 51.75$ | A1 | cso 51.75 |